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anonymous

  • one year ago

Suppose ΔABC is an acute triangle. Let the bases of the altitudes at B and C be E and F, respectively. Prove ΔABC and ΔAEF are similar

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  1. anonymous
    • one year ago
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    I typed the question verbatim, but it sounds odd that E and F would be bases. I would think they would be points. Regardless of that, still am unsure of how to do the problem.

  2. ganeshie8
    • one year ago
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    |dw:1437110964213:dw|

  3. anonymous
    • one year ago
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    Oh, okay, so they really are supposed to be points E and F. Maybe I drew my picture wrong but, at least the way yours is drawn, you could maybe show FE is parallel to BC?

  4. ganeshie8
    • one year ago
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    That is equivalent, but there is a simpler way here

  5. anonymous
    • one year ago
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    Oh, thats equivalent to saying theyre similar? Saying those lines are parallel?

  6. ganeshie8
    • one year ago
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    First establish trivially that \(\triangle OFB\sim\triangle OEC\) by \(AA\) |dw:1437112233723:dw|

  7. anonymous
    • one year ago
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    Right, made it that far at least xD

  8. ganeshie8
    • one year ago
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    Next show that \(\triangle AFC\sim AEB\) by AA |dw:1437112450857:dw| actually we don't need the first step :)

  9. ganeshie8
    • one year ago
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    setup a proportion and we're done!

  10. anonymous
    • one year ago
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    Alright, I have that similarity. After that, maybe because I was being confused by my picture, I wasnt sure. And we're setting up a proportion based on the sides of triangle AFC and triangle AEB?

  11. ganeshie8
    • one year ago
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    Yes : \[\dfrac{AF}{AE}=\dfrac{AC}{AB}\] together with \(\angle A\cong \angle A\) by reflexive property yield \(\triangle ABC\sim\triangle AEF\) by \(SAS\) similarity

  12. anonymous
    • one year ago
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    Well, we didn't really talk about a lot of the high school geometry postulates and such. It seems like the professor just assumes that what we are given implies everything else. Maybe what we were given implies SAS similarity? This is all we were given about similar triangles: "ΔABC ~ ΔA'B'C' iff \(\frac{|A'B'|}{|AB|} = \frac{|A'C'|}{|AC|} = \frac{|B'C'|}{|BC|}\)" Or "ΔABC ~ ΔA'B'C' iff the angle at A = angle at A', angle at B = angle at B', and angle at C = angle at C'"

  13. ganeshie8
    • one year ago
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    "ΔABC ~ ΔA'B'C' iff the angle at A = angle at A', angle at B = angle at B', and angle at C = angle at C'" This is equivalent to AA similarity because the third angle is determined by triangle sum property once the first two angles are known.

  14. ganeshie8
    • one year ago
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    "ΔABC ~ ΔA'B'C' iff \(\frac{|A'B'|}{|AB|} = \frac{|A'C'|}{|AC|} = \frac{|B'C'|}{|BC|}\)" This is SSS similarity, lets see if we can jump to this from SAS

  15. ganeshie8
    • one year ago
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    here it is, pls scroll directly to SAS similarity http://ceemrr.com/Geometry1/SSSandSASsimilarity/SSSandSASsimilarity_print.html they're proving SAS similarity using AA

  16. anonymous
    • one year ago
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    Alrighty, Ill take a look at that and make sure I see the SAS. Thanks again :)

  17. ganeshie8
    • one year ago
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    This is analogous to SAS congruence : two triangles are congruent if the corresponding pairs of two adjacent sides and the included angle are congruent

  18. ganeshie8
    • one year ago
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    SAS similarity says : two triangles are "similar" if the corresponding pairs of two adjacent sides are "proportional" and the included angles are congruent.

  19. anonymous
    • one year ago
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    In the link, the statement they give for SAS says "If \(\frac{AB}{DE} = \frac{AC}{DF}\) and \(∠A\) congruent to \(∠B\)..." Yet in their diagram, it looks as if it would need to say that \(∠A\) congruent to \(∠D\)

  20. ganeshie8
    • one year ago
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    Ahh, I overlooked that. thats definitely a typo yeah

  21. ganeshie8
    • one year ago
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    |dw:1437114236422:dw|

  22. anonymous
    • one year ago
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    Okay, cool. I gotta learn to use the latex and drawing tools and such like you know how to, haha.

  23. ganeshie8
    • one year ago
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    I'm using a small chrome extension that gives plotting/graphing like features to the drawing tool

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